Question

A student solved the equation sin2x/cos x = 2, 0 ≤ x ≤ pi, and got pi/2. What was the student's error?

Answer 1

See below

Explanation:

Since
`cos((\pi)/(2))=0`, the denominator will be 0, and the result would be indeterminate.

Do not read on if you do not understand Calculus

However, notice that
`\lim_{x \to (\pi)/(2)} (sin(2x))/(cos(x))=2` by L'Hopital's Rule:

`\lim_{x \to (\pi)/(2)} (sin(2x))/(cos(x))\\\\(2cos(2x))/(-sin(x))\\ \\(2cos(2((\pi)/(2))))/(-sin((\pi)/(2) ))\\\\(2cos(\pi))/(-1)\\ \\(-2)/(-1)\\ \\2`

Answer 2

Explanation:

Albeit cosine is undefined at , this discontinuity is evitable by canceling terms. The expression is equivalent to , which is continuous for all . According to this result, the student did not make a mistake. Answer was correct.